Over the weekend, we placed an order and drove up to the city to pick it up (their warehouse is open for pickups from 12-1pm, Monday - Saturday). They filled our little cooler and with ice and seafood. With the pickup discount, we got:
- 1 dozen Hood Canal Bay oysters ($16)
- 1.5 lb Hawaiian pole-and-line albacore tuna ($20)
- 1 lb Prince Edward Island mussels ($6)
The oysters were amazing. Difficult to open (we both ended up stabbing ourselves with a 3" paring knife), but delicious.
It turns out we were storing them a bit colder than optimal (ideal is 35-40 °F, with adequate oxygen supply), but we ate them over the course of 36 hours without ill effect.
At only $6, I think the mussels were best deal of all. We ate a few raw (I don't recommend it - they're kinda bitter) and boiled the rest to make a sauce for linguine (olive oil, white wine, mussels, bell pepper, tomato). All the mussels were alive, and there was enough for 2 plates like this:
We then had seared tuna for lunch the next day. The searing adds a welcome complexity of taste to the surface, while the bulk maintains the texture of the raw fish. We want the surface temperature to reach 150°C for the Maillard reaction, but want to minimize the region that exceeds 60°C (turns opaque and flaky). So if raw albacore tuna has a thermal diffusivity of 0.15 mm2/s and we start at an initial condition of 0°C, this is starting to sound like a parabolic PDE...
The boundary condition is problematic. Making it isothermal with the pan defeats the purpose of our modeling; perhaps constant heat flux is more appropriate. Furthermore, the meat's heat capacity becomes ∞ for a while when it reaches 100°C, but we can approximate this effect by adding fMΔHvap/CP to the goal temperature, where fM is the tuna's moisture content (63%), ΔHvap is water's heat of vaporization (2257 J/g), and CP is the specific heat of tuna (3.1 J/g K). This gives us a target surface temp of 610°C...
Okay, looks like ΔHvap is a bigger player than I initially thought. Within the fish, conduction would be best modeled with an isothermal boundary condition at 100°C. Using the semi-infinite slab solutions (which is actually just an ODE after nondimensionalization), we get T(x,t) = Tsurfaceerfc(x/√4αt). We're interested in when/where T = 60°C, i.e. x/√4αt = erfc-1(60/100) = 0.37. For x in mm and t in s, this comes out to x = 0.29√t. So that ~1 mm "cooked" layer in the picture would have taken 12 sec.
The next question is to find how long it takes for the fish surface to reach 150°C. Unfortunately this requires more information, including the heat flux into the surface and some sort of diffusion length for the movement of moisture from the bulk into the rapidly-dehydrating surface. These are the primary factors in determining how long it will take to brown the surface.
As a side note, I have got to figure out how to put LaTeX into blog posts; formatting formulae in HTML is terribly unwieldy.
Anyways, we still had leftover fish after the seared tuna lunch, so we've salted the rest in much the same way as we did before.
The boundary condition is problematic. Making it isothermal with the pan defeats the purpose of our modeling; perhaps constant heat flux is more appropriate. Furthermore, the meat's heat capacity becomes ∞ for a while when it reaches 100°C, but we can approximate this effect by adding fMΔHvap/CP to the goal temperature, where fM is the tuna's moisture content (63%), ΔHvap is water's heat of vaporization (2257 J/g), and CP is the specific heat of tuna (3.1 J/g K). This gives us a target surface temp of 610°C...
Okay, looks like ΔHvap is a bigger player than I initially thought. Within the fish, conduction would be best modeled with an isothermal boundary condition at 100°C. Using the semi-infinite slab solutions (which is actually just an ODE after nondimensionalization), we get T(x,t) = Tsurfaceerfc(x/√4αt). We're interested in when/where T = 60°C, i.e. x/√4αt = erfc-1(60/100) = 0.37. For x in mm and t in s, this comes out to x = 0.29√t. So that ~1 mm "cooked" layer in the picture would have taken 12 sec.
The next question is to find how long it takes for the fish surface to reach 150°C. Unfortunately this requires more information, including the heat flux into the surface and some sort of diffusion length for the movement of moisture from the bulk into the rapidly-dehydrating surface. These are the primary factors in determining how long it will take to brown the surface.
As a side note, I have got to figure out how to put LaTeX into blog posts; formatting formulae in HTML is terribly unwieldy.
Anyways, we still had leftover fish after the seared tuna lunch, so we've salted the rest in much the same way as we did before.
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